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ECE 472/572 - Digital Image Processing

Lecture 6 – Geometric and Radiometric Transformation 09/27/11

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Roadmap

¬  Introduction –  Image format (vector vs. bitmap) –  IP vs. CV vs. CG –  HLIP vs. LLIP –  Image acquisition

¬  Perception –  Structure of human eye –  Brightness adaptation and Discrimination –  Image resolution

¬  Image enhancement –  Enhancement vs. restoration –  Spatial domain methods

•  Point-based methods –  Log trans. vs. Power-law –  Contrast stretching vs. HE –  Gray-level vs. Bit plane slicing –  Image averaging (principle)

•  Mask-based methods - spatial filter –  Smoothing vs. Sharpening filter –  Linear vs. Non-linear filter –  Smoothing (average vs. Gaussian vs. median) –  Sharpening (UM vs. 1st vs. 2nd derivatives)

–  Frequency domain methods •  Understanding Fourier transform •  Implementation in the frequency domain •  Low-pass filters vs. high-pass filters vs. homomorphic filter

¬  Geometric correction –  Affine vs. Perspective

transformation •  Homogeneous coordinates •  Inverse vs. forward transform •  Composite transformation

–  General transformation •  Model distortion with polynomial •  Least square solution

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Questions

¬ Affine transformation vs. Perspective transformation

¬ Forward transformation vs. Inverse transformation

¬ Composite transformation vs. Sequential transformation

¬ Homogeneous coordinate ¬ General geometric transformations

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Usage

¬ Image correction ¬ Color interpolation ¬ Forensic analysis ¬ Entertainment effect

http://w3.impa.br/~morph/ http://www.mpi-sb.mpg.de/resources/FAM/demos.html

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Affine transformations

¬ Preserve lines and parallel lines ¬ Homogeneous coordinates ¬ General form

¬ Special matrices –  R: rotation, S: scaling, T: translation, H: shear

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R =

cosθ −sinθ 0sinθ cosθ 00 0 1

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,S =

sx 0 00 sy 00 0 1

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,T =

1 0 tx0 1 ty0 0 1

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,H =

1 hx 0hy 1 00 0 1

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uv1

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=

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Composite vs. Sequential transformation

Original Image (f(x,y))

R H T S

Transformed Image (g(u, v))

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uv1

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' ' '

= S ⋅T ⋅H ⋅ R ⋅xy1

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RHTSC ⋅⋅⋅=

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Forward vs. Inverse transforms

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uv1

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= C ⋅xy1

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Forward transform

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C−1

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Inverse transform

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Examples - Shear

hy = 0.2 hx = 0.2 hx = hy = 0.2

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Examples – Translation + Rotation

theta = PI/4 theta = PI/4 tx = -140, ty = 60

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Perspective transformation

¬ Preserve parallel lines only when they are parallel to the projection plane. Otherwise, lines converge to a vanishing point

¬ General form

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" u " v " w

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xy1

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,u =" u " w ,v =

" v " w

0,0 3231 ≠≠ aa

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Determine the coefficients

8 unknowns, 4-point least squares

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,u =" u " w ,v =

" v " w

(0,0) (0,255)

(255,0) (255,255)

(0,0) (0,255)

(255,0)

(511,511)

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Example - PT

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General approaches

¬ Find tiepoints ¬ Spatial transformation

9/27/11 15

x-ray sensitive scintillator

fiber optics

CCD array 1242 x 1152

misalignment greater than 50 micron

Example – CCD butting

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9/27/11 16

Sources of distortions

¬ defects in the production of fiber-optic tapers

¬ imperfect compression and cutting ¬ different light transfer efficiency across the

whole surface

9/27/11 17

),(),( yxvuP =

),(),( yxvuT =

approximation

interpolation

),( yx ),( vu

control point

map exactly

map close

Geometric correction

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Spatial transformation

¬ Bilinear equation

¬ n-th degree polynomial

¬ Use information from tiepoints to solve coefficients –  Exact solution –  Least square solution

€

ˆ x = r u,v( ) = a1u + a2v + a3uv + a4

ˆ y = s u,v( ) = b1u + b2v + b3uv + b4

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ˆ x iˆ y i

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Px (ui,vi)Py (ui,vi)"

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akrsuirvi

s

bkrsuirvi

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r+s= k∑

k= 0

d

∑

∑−

=

−+−=1

0

22, ])ˆ()ˆ[(minm

iiiiiba yyxxε

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How is it applied?

¬  Step 1: Choose a set of tie points –  (xi,yi): coordinates of tie points in

the original (or distorted) image –  (ui,vi): coordinates of tie points in

the corrected image ¬  Step 2: Decide on which degree of

polynomial to use to model the inverse of the distortion, e.g.,

¬  Step 3: Solve the coefficients of the polynomial using least-squares approach

¬  Step 4: Use the derived polynomial model to correct the entire original image

€

X = x0 x1 x24[ ]T

Y = y0 y1 y24[ ]T

€

ˆ x = r u,v( ) = a1u + a2v + a3uv + a4

ˆ y = s u,v( ) = b1u + b2v + b3uv + b4

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W =

u0 v0 u0v0 1u1 v1 u1v1 1 u24 v24 u24v24 1

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A = a1 a2 a3 a4[ ]T

B = b1 b2 b3 b4[ ]T

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A =W −1X,B =W −1YW −1 = (W TW )−1W T

For each (u,v) in the corrected image, find the corresponding (x,y) in the original image and use its intensity as the intensity at (u,v).

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Example – Geometric correction

¬ Geometric correction of images from butted CCD arrays

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Example - Color correction

Tiepoints are colors (R, G, B), instead of spatial coordinates

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Example - Image warping

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Image warping

¬ Two-pass mesh warping by Douglas Smythe

¬ Reference: G. Wolberg, Digital Image Warping, 1990

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Example 1

From Joey Howell and Cory McKay, ECE472, Fall 2000

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Example 2

From Adam Miller, Truman Bonds, Randal Waldrop, ECE472, Fall 2000